Question

If one zero of a polynomial p(x) = (k? + 4).x2 + 13x +
4k is reciprocal of the other, then prove that k = 2.​

Answer 1

Step-by-step explanation:

Let One Zero Be 'p' and other be 1/p

product of roots of quadratic equation=c/a

p*1/p=(4k)/(k^2+4)

1=4k/(k^2+4)

 after cross multiplication

 k^2+4=4k

k^2-4k+4=0

k^2-2k-2k+4=0

k(k-2)-2(k-2)=0

(k-2)(k-2)=0

(k-2)^2=0

k-2=0

k=2

HENCE PROVED

Answer 2

$\fbox\pink{Solution}$

(k² + 4)x² + 13x + 4k = 0

The roots of this polynomial are reciprocal of each other.

Let one root be a .

So, the other root will be $\tt\purple{\dfrac{1}{a}}$

From the equation, the product of root is $\tt \purple{ \dfrac{4k}{k {}^{2} + 4 } }$

So,

$\tt \purple{ a \dfrac{1}{a} = \dfrac{4k}{k {}^{2} + 4 } } \\ \\ \tt \: \purple{ 1 = \frac{4k}{k {}^{2} + 4 } } \\ \\ \tt \purple{\twoheadrightarrow \: k {}^{2} + 4 - 4k = 0 } \\ \\ \tt \purple{\twoheadrightarrow \: (k - 2) {}^{2} = 0} \\ \\ \tt\purple{\twoheadrightarrow \: k = 2}$

$\boxed {\boxed{ \tt \pink H \purple e \pink n \purple c \pink e, \: \: \: \purple p \pink r \purple o \pink v\purple e \pink d \: }}$

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$\fbox\pink{ @ EuphoriaBunny}$